>> f={(x,x^2+1) | x is in R)}
Took liberty changing x2 to x-squared (is this right?)
This looks like a mapping of all x in R.
Given any point x in the x-axis (in R), then the corresponding point along the y-axis (also in R) is x-squared.
So, the function f is just a parabola: f(x) = x^2+1

From that blog and the previous links, I am inclined to take issue with your notation: f={(x^2+1) | x is in R)}
It seems to be infering a mapping from R -> R as x -> x^2 + 1, but I think you need to be more explicit, as in:

f: R -> R, defined by { x -> (x^2+1) | x is in R) }
or,
f: R -> R, defined by x -> x^2 + 1 or equivalently, f(x) = x^2 + 1
but, notationally, f(x), may be a little misleading, since, on one hand, f(x) is just a number, given a number x; yet, we have learned it in H.S. algebra that f(x) is a function. The blog tries to help clarify these points and relates algebra to set theory.

Paraphrasing from this blog and applying to your problem..
Given a function f: R -> R , we can define a natural notion of the graph of that function. It is the set of all points ( x, f(x) ) where x is an element in R. To put it another way, it is the set of all points (x, y) that are in R x R such that y = f(x) = (x^2+1).

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This Article is a follow-up to my Mappit! Addin Article (http://www.experts-exchange.com/A_2613.html), it was inspired by an email posting I made to EUSPRIG (http://www.eusprig.org/index.htm),
I will briefly cover:
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This web page has appeared at Google. It's definitely worth considering!
https://www.google.com/about/careers/students/guide-to-technical-development.html
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